Three vectors →P,→Q and →R are shown in the figure. Let S be any point on the vector →R. The distance between the points P and S is b|→R|. The general relation among vectors →P,→Q and →S is
A
→S=(1−b)→P+b2→Q
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B
→S=(1−b2)→P+b→Q
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C
→S=(1−b)→P+b→Q
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D
→S=(b−1)→P+b→Q
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Solution
The correct option is D→S=(1−b)→P+b→Q
From triangular law of vector addition, we get OP+PS=OS